Understanding the morphism of schemes involving nilpotents induced by: $\mathbb C[x] \to \mathbb C[x]/(x^2)$ where $x \mapsto ax \pmod {x^2}$

Question

From Vakil's FOAG:

The following imprecise exercise will give you some sense of how to visualize maps of schemes when nilpotents are involved. Suppose $a \in \mathbb C$. Consider the map of rings $\mathbb C[x] \to \mathbb C[x]/(x^2)$ given by $x \mapsto ax$. Recall that $\operatorname{Spec} \mathbb C[x]/(x^2)$ may be pictured as a point with a tangent vector (§4.2). How would you picture this map if $a \ne 0$? How does your picture change if $a=0$?

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When $a \ne 0$, I believe the ring map $\mathbb C[x] \to \mathbb C[x]/(x^2)$ given by $x \to ax$ induces the morphism of schemes $\operatorname{Spec} \mathbb C[x]/(x^2) \to \operatorname{Spec} \mathbb C[x]$ given by $(x)/(x^2) \mapsto (x)$. I think this is true because if $a \ne 0$, then $(x)/(x^2)=(ax)/(x^2)$ and the preimage of $(ax)/(x^2)$ is $(x)$.

When $a=0$, then we also have morphism of schemes $\operatorname{Spec} \mathbb C[x]/(x^2) \to \operatorname{Spec} \mathbb C[x]$ given by $(x)/(x^2) \mapsto (x)$ because the preimage of $(x)/(x^2)$ is the set of all polynomials with zero constant term, i.e., the prime ideal $(x)$.

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Is the above correct? I am still not sure how to visualize this morphism.

What is this exercise trying to get across?

Answer 1

Pointwise, you are correct.

This is my rookie view of why he says that the tangent vector should be “crushed”. Suppose you have a function $f(x)$ on $\operatorname{Spec} \mathbb C[x].$ Then under the map $\operatorname{Spec} \mathbb C[\epsilon]/(\epsilon^2) \to \operatorname{Spec} \mathbb C[x]$ it pulls back to $f(a\epsilon).$ Now, for $a \not =0$ having the function $\epsilon \mapsto f(a\epsilon)$ in $\operatorname{Spec} \mathbb C[\epsilon]/(\epsilon^2)$ means knowing the value of $f$ plus its derivative. Thus, the tangent vector accounts for the fact that you can take the derivative along it. However, when $a=0$, $f(x)$ pulls back to $f(0)$ and your information about the derivative is lost.

Answer 2

The map $$f_a:X=\operatorname{Spec} \mathbb{C}[x]/(x^2)\to Y =\operatorname{Spec} \mathbb{C}[x] $$ induced by $$ax\leftarrow x.$$ By definition of tangent spaces, $T_OX=\Bigg(\dfrac{(\bar{x})}{(\bar{x})^2}\Bigg)^{\vee}$ and $T_OY=\Bigg(\dfrac{(x)}{(x)^2}\Bigg)^{\vee}$ where $\bar{x}$ is the image of $x$ in $\mathbb{C}[x]/(x^2)$. The map $$df_a:T_OX\to T_OY$$ of the tangent spaces at the origin $O$ is determined by the image of $\bar{x}^{\vee}$ (the $\mathbb{C}$-linear map taking $\bar{x}\mapsto 1$)

given by $$\bar{x}^{\vee}\mapsto a\cdot x^{\vee}.$$

Hence $df_a$ scales the tangent vector at the origin $O$ by scalar $a$.

In particular, for $a= 0$, $df_a$ crushes the tangent vector to $0$.